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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: LaçinCanAtış - Temmuz 23, 2016, 11:14:07 ös
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$a,b,c\ge 0$ ve $a+b+c\le 1$
ise
$$16\left(\sum a^2b^2\right)+11abc\le 1$$
olduğunu kanıtlayınız.
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$0\le c\le 1-\left(a+b\right)$ dir.O halde
$$16\left(\sum a^2b^2\:\right)+11abc\le 16\left(\left(ab\right)^2+c^{2\:}\left(a^2+b^2\right)\right)+11abc\\ $$ olur.
$s=a+b,t=ab$
$f\left( x \right) =x^{ 2 }-sx+t\: \quad a,b\in \left[ 0,1 \right] \Longleftrightarrow \left\{ f\left( 0 \right) \ge 0,f\left( 1 \right) \ge 0 \right\} \quad \Longrightarrow 0\le t\le \frac { s^{ 2 } }{ 4 } $.
$f\left( x \right) =x^{ 2 }-sx+t\: \quad a,b\in \left[ 0,1 \right] \Longleftrightarrow \left\{ f\left( 0 \right) \ge 0,f\left( 1 \right) \ge 0 \right\} \quad \Longrightarrow 0\le t\le \frac { s^{ 2 } }{ 4 } \\ 16\left( \sum a^{ 2 }b^{ 2 }\: \right) +11abc\le 16\left( \left( ab \right) ^{ 2 }+c^{ 2\: }\left( a^{ 2 }+b^{ 2 } \right) \right) +11abc\\ \le 16\left( t^{ 2 }+\left( 1-s \right) ^{ 2 }\left( s^{ 2 }-2t \right) \right) +11t\left( 1-s \right) =16t^{ 2\: }+\left( 1-s \right) \left( 32s-21 \right) t+16s^{ 2 }\left( 1-s \right) ^{ 2 }\\ \left( 1-s \right) \left( 32s-21 \right) \ge 0\quad dır.\\ \le 16\left(\frac{s^2}{4}\right)^2+\frac{\left(1-s\right)\left(32s-21\right)s^2}{4}+16s^2\left(1-s\right)^2=\frac{1}{4}\left(36s^4-75s^2+43s^2\right)=\frac{1}{4}\left(4s+1\right)\left(s-1\right)\left(3s-2\right)^2+1\le 1$